On Fri, 5 Oct 2001 06:00:07 +1000 (EST), Damian Conway wrote:
> > $foo + $bar
> >
> > will call $foo's overloaded add if it has one no matter where $foo's used.
>
>Sorry. I should have been clearer.
>
>Dan is, of course, correct. Overloaded operators that are class methods
>will tag along with their objects as they currently do (as indeed, *all*
>methods currently do).
What do you mean, "class methods"? Here, "+" isn't a class method, but
an object method. Well, assuming $foo is an object, of course.
--
Bart.
